Problem: Let $h(x)=\dfrac{x^2-49}{x+7}$ when $x\neq -7$. $h$ is continuous for all real numbers. Find $h(-7)$. Choose 1 answer: Choose 1 answer: (Choice A) A $7$ (Choice B) B $14$ (Choice C) C $-7$ (Choice D) D $-14$
$\dfrac{x^2-49}{x+7}$ is continuous for all real numbers other than $x=-7$ which means $h$ is continuous for all real numbers other than $x=-7$. In order for $h$ to also be continuous at $x=-7$, the following equality must hold: $\lim_{x\to -7}h(x)=h(-7)$ We will obtain the above equality by letting $h(-7)=\lim_{x\to -7}h(x)$. So let's find $\lim_{x\to -7}h(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to -7}h(x) \\\\ &=\lim_{x\to -7}\dfrac{x^2-49}{x+7} \gray{\text{This is the rule for }x\neq -7} \\\\ &=\lim_{x\to -7}\dfrac{\cancel{(x+7)}(x-7)}{\cancel{(x+7)}} \gray{\text{Factor}} \\\\ &=\lim_{x\to -7}{(x-7)} \gray{\text{Cancel common factors}} \\\\ &\text{(This is allowed because }x\neq -7) \\\\ &=(-7)-7 \gray{\text{Direct substitution}} \\\\ &=-14 \end{aligned}$ We obtained that if we set $h(-7)=-14$, then $\lim_{x\to -7}h(x)=h(-7)$, which makes $h$ continuous at $x=-7$. Since we already saw that $h$ is continuous for any other real number, we can determine that it's continuous for all real numbers. In conclusion, $h(-7)=-14$.